我最近发布了另一个与搜索引擎相关的帖子,我正在基于Kevin Yank的"建立自己的DB驱动网站等"。搜索引擎正在工作,但是有一些错误,我想修复,不知道如何。在这篇文章中,我只关注其中的一个,这样就不会让人太困惑了。
在数据库中,有一个表用于笑话(称为"joke"),另一个表用于主题(称为"theme")。这两个表由另一个名为"joketheme"的表连接。每个笑话应该能够有一个以上的主题,我希望结果列出所有的主题,而不重复的条目。到目前为止,我还没能做到这一点。事实上,我不确定如何在MySQL中为一个笑话指定2个主题。 table 1: joke
id~~~joketext~~~other data
1~~~joke1~~~
2~~~joke2~~~
3~~~joke3~~~
table 2: theme
id~~~name
1~~Knock knock
2~~Lawyers
table 3: joketheme
jokeid~~~themeid
1~~~~~~1
1~~~~~~2
2~~~~~~2
3~~~~~~1
你知道我需要改变的代码(或在MySQL)在每个结果中列出多个主题(如果该条目有多个主题?)?
这是我的代码在搜索结果页面。提前感谢您的任何帮助!:
<?php
$dbcnx = @mysql_connect('localhost', 'root', 'password');
if (!$dbcnx) {
exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}
if (!@mysql_select_db('ijdb')) {
exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
}
$authors = @mysql_query('SELECT id, name FROM author');
if (!$authors) {
exit('<p>Unable to obtain author list from the database.</p>');
}
$cats = @mysql_query('SELECT id, name FROM category');
if (!$cats) {
exit( '<p>Unable to obtain category list from the database.</p>');
}
$themes = @mysql_query('SELECT id, name FROM theme');
if (!$themes) {
exit( '<p>Unable to obtain category list from the database.</p>');
}
$geofoci = @mysql_query('SELECT id, name FROM geofocus');
if (!$geofoci) {
exit( '<p>Unable to obtain category list from the database.</p>');
}
?>
搜索表单:
<form class="searchField" name="input" action="fundfetch_search.php" method="post">
<ul>
<li>
<label>Search by keyword:</label>
<input type="text" name="searchtext" class="styleSearchbox" placeholder="By keyword" value="<?php echo $_POST['searchtext']; ?>">
</li>
<li>
<label>OR by the following: </label>
<label><select name="aid" size="1" class="styleDropdown">
<option selected value="">Any Author</option>
<?php
while ($author = mysql_fetch_array($authors)) {
$aid = $author['id'];
$aname = htmlspecialchars($author['name']);
echo "<option value='$aid'>$aname</option>'n";
}
?>
</select></label>
</li>
<li>
<label><select name="cid" size="1" class="styleDropdown">
<option selected value="">Any Category</option>
<?php
while ($cat = mysql_fetch_array($cats)) {
$cid = $cat['id'];
$cname = htmlspecialchars($cat['name']);
echo "<option value='$cid'>$cname</option>'n";
}
?>
</select></label>
</li>
<li>
<label><select name="tid" size="1" class="styleDropdown">
<option selected value="">Any Theme</option>
<?php
while ($theme = mysql_fetch_array($themes)) {
$tid = $theme['id'];
$tname = htmlspecialchars($theme['name']);
echo "<option value='$tid'>$tname</option>'n";
}
?>
</select></label>
</li>
<li>
<label><select name="gfid" size="1" class="styleDropdown">
<option selected value="">Any Region</option>
<?php
while ($geofocus = mysql_fetch_array($geofoci)) {
$gfid = $geofocus['id'];
$gfname = htmlspecialchars($geofocus['name']);
echo "<option value='$gfid'>$gfname</option>'n";
}
?>
</select></label>
</li>
<li style="visibility:hidden"><a href="../FUNDER.COM website/searchfilteroption">Closing</a></li>
<li><input type="submit" value="Search" class="searchButton"></li>
</ul>
</form>
查询: <?php
$dbcnx = @mysql_connect('localhost', 'root', 'password');
if (!$dbcnx) {
exit('<p>Unable to connect to the ' . 'database server at this time.</p>');
}
if (!@mysql_select_db('ijdb')) {
exit('<p>Unable to locate the joke ' . 'database at this time.</p>');
}
$select = 'SELECT DISTINCT joke.id, joke.joketext, joke.jokedate,
author.id AS author_id, author.name AS author_name,
jokecategory.jokeid AS cat_jokeid, jokecategory.categoryid AS joke_catid, category.id AS cat_id, category.name as cat_name,
joketheme.jokeid AS theme_jokeid, joketheme.themeid AS joke_themeid, theme.id AS theme_id, theme.name AS theme_name,
jokegeofocus.jokeid AS geofocus_jokeid, jokegeofocus.geofocusid AS joke_geofocusid, geofocus.id AS geofocus_id, geofocus.name AS geofocus_name';
$from = ' FROM joke, author, jokecategory, category, joketheme, theme, jokegeofocus, geofocus';
$where = ' WHERE joke.authorid = author.id AND joke.id = jokecategory.jokeid AND jokecategory.categoryid = category.id AND joke.id = joketheme.jokeid AND joketheme.themeid = theme.id AND joke.id = jokegeofocus.jokeid AND jokegeofocus.geofocusid = geofocus.id';
$in = ' ORDER BY jokedate DESC';
$aid = $_POST['aid'];
if ($aid != '') { // An author is selected
$where .= " AND authorid='$aid'";
}
$cid = $_POST['cid'];
if ($cid != '') { // A category is selected
$from .= ''; // usually written as ' ,tablename'
$where .= " AND joke.id=jokecategory.jokeid AND categoryid='$cid'";
}
$tid = $_POST['tid'];
if ($tid != '') { // A theme is selected
$from .= '';
$where .= " AND joke.id=joketheme.jokeid AND themeid='$tid'";
}
$gfid = $_POST['gfid'];
if ($gfid != '') { // A region is selected
$from .= '';
$where .= " AND joke.id=jokegeofocus.jokeid AND geofocusid='$gfid'";
}
$searchtext = $_POST['searchtext'];
if ($searchtext != '') { // Some search text was specified
$where .= " AND keywords LIKE '%$searchtext%'";
}
?>
结果 <?php
$jokes = @mysql_query($select . $from . $where . $in);
if (!$jokes) {
echo '</table>'; exit('<p>Error retrieving jokes from database!<br />'.
'Error: ' . mysql_error() . '</p>');
}
$numrows = mysql_num_rows($jokes);
if ($numrows>0){
while ($joke = mysql_fetch_array($jokes)) {
$id = $joke['id'];
$joketext = htmlspecialchars($joke['joketext']);
$jokedate = htmlspecialchars($joke['jokedate']);
$aname = htmlspecialchars($joke['author_name']);
$category = htmlspecialchars($joke['cat_name']);
$theme = htmlspecialchars($joke['theme_name']);
$geofocus = htmlspecialchars($joke['geofocus_name']);
$position = 200;
$post = substr($joketext, 0, $position);
echo "<li id='"jump'">
<article class='"entry'">
<header>
<h3 class='"entry-title'"><a href=''>$aname</a></h3>
</header>
<div class='"entry-content'">
<p>$post...</p>
</div>
<div class ='"entry-attributes'">
<p>> Category: $category</p>
<p> > Theme(s): $theme</p>
<p> > Region(s) of focus: $geofocus</p>
</div>
<footer class='"entry-info'">
<abbr class='"published'">$jokedate</abbr>
</footer>
</article>
</li>";
}
}
else
echo "Sorry, no results were found. Please change your search parameters and try again!";
?>
听起来您需要对SQL查询及其工作原理进行一些研究。如果想要单个笑话的主题列表,可以执行如下简单查询:
SELECT DISTINCT(theme.name) FROM theme
INNER JOIN joketheme ON joketheme.theme_id = theme.id
INNER JOIN joke ON joke.id = joketheme.joke_id
WHERE joke.id = {insert joke id}
我们在这里做的叫做join。我们将三个表连接在一起,根据所有表中通用的id匹配结果。这些被称为钥匙。基本上,我们正在尝试从主题表中选择与特定笑话主题匹配的所有主题。
查询的DISTINCT部分确保我们只得到唯一的结果。
我建议你看看一些SQL教程,了解一些使用查询获取不同数据集的不同方法。知道这些东西真的很有用,可以为你节省大量的代码。
另外,就可读性而言,我建议在连接表的表名之间加上下划线(例如joke_theme),或者采用其他类似的约定。使它更容易阅读和分辨哪个是连接表,哪个是普通表。
祝你好运